# The equation a^3+b^3+c^3=2008 has a solution in which a,b,and c are distinct even positive integers. find a+b+c?

Jun 18, 2018

The answer is $= 22$

#### Explanation:

The equation is

${a}^{3} + {b}^{3} + {c}^{3} = 2008$

Since $a , b , c \in \mathbb{N}$ and are even

Therefore,

$a = 2 p$

$b = 2 q$

$c = 2 r$

Therefore,

${\left(2 p\right)}^{3} + {\left(2 q\right)}^{3} + {\left(2 r\right)}^{3} = 2008$

$\implies$, $8 {p}^{3} + 8 {q}^{3} + 8 {r}^{3} = 2008$

$\implies$, ${p}^{3} + {q}^{3} + {r}^{3} = \frac{2008}{8} = 251$

$\implies$, ${p}^{3} + {q}^{3} + {r}^{3} = 251 = {6.3}^{3}$

Therefore,

$p$, $q$ and $r$ are $\le 6$

Let $r = 6$

Then

${p}^{3} + {q}^{3} = 251 - {6}^{3} = 35$

${p}^{3} + {q}^{3} = {3.27}^{3}$

Therefore,

$p$ and $q$ are $\le 3$

Let $q = 3$

${p}^{3} = 35 - {3}^{3} = 35 - 27 = 8$

$\implies$, $p = 2$

Finally

$\left\{\begin{matrix}a = 4 \\ b = 6 \\ q = 12\end{matrix}\right.$

$\implies$, $a + b + c = 4 + 6 + 12 = 22$